Question

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at constant volume, adding 1.36 * 104 J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

Answer 1

Answer:

a)  $T_b=590.775k$

b)  $W_t=1.08*10^4J$

d)  $Q=3.778*10^4J$

d)  $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

                      $t = 20+273$

                     $t = 293k$

Initial heat $Q=1.36 * 10^4 J$

a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

  Where

  $C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

 $PV=nRT$

For v double

 $T_c=2*590.775k$

 $T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$